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Proof of Gauss's Law 
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#19
Sep306, 10:00 AM

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A field like [itex]\vec F(\vec r)=K \hat r/r^3[/itex] is radial, but not divergenceless. That is a property unique to the inverse square relationship. In fact, an inverse square field is the only radial field that is divergenceless. 


#20
Sep306, 10:40 AM

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#21
Sep306, 04:53 PM

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[tex] \vec{r} = r\,\hat{r} + \arctan(\frac{y}{x})\,\hat{\phi} + \arcsin(\frac{\sqrt{x^2+y^2}}{\sqrt{x^2+y^2+z^2}})\,\hat{\theta}[/tex] How is this more easier? 


#22
Sep306, 11:13 PM

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No way. In spherical coordinates the field is simply [itex]\hat r/r^2[/itex]. There is no phi or theta dependence in there, since it's radial remember? All you have to do is lookup the divergence in spherical coordinates and do a simply derivative.



#23
Sep406, 12:43 AM

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#24
Sep406, 02:14 AM

P: 212

If you happen to have an E&M book handy, there's probably some derivation of the integral form of Gauss's law from the generalized coulomb's law, and then the derivation of the differential form from the integral form, which is fairly trivial using the divergence theorem.
There's a fairly rigorous derivation in Classical Electrodynamics, by Jackson p2729, and a somewhat lessthanrigorous (but quite descriptive) derivation in Introduction to Electrodynamics, by Grffiths, p6570 


#25
Sep406, 08:53 AM

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#26
Sep506, 01:13 PM

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#27
Aug910, 08:35 AM

P: 2

Hey i have a new proof of gauss law.now i am listing it below
let there be an arbitaraly shaped body.let there be a charge Q,e='epsilon not',E=electric field,dS=small surface element. let the body be consisting of many small surface elements dS_{1},dS_{2}...... let the individual surfaces be so small that radius here are constant and be r_{1},r_{2}......... respectively therefore surface integral of E.dS=E_{1}.dS_{1}+E_{2}.dS_{2}+..... since theta =0 therefore E.dS=E*dS therefore putting values Q*dS_{1}/(4*pi*e*r_{1}^{2})+Q*dS_{2}/(4*pi*e*r_{2}^{2})+.... d(theta_{1})=dS_{1}/dr_{1}^{2},d(theta_{2})=dS_{2}/dr_{2}^{2},etc (from definition of solid angle) taking everything in common net elctric flux=Q/(4*pi*e)(dS_{1}/dr_{1}^{2}+dS_{2}/dr_{2}^{2}+...) =Q/(4*pi*e)(d(theta_{1})+d(theta_{2})+....) =Q/(4*pi*e)(4*pi)=Q/(e) (since sum of all solid angle around the charge is 4*pi) 


#28
Aug910, 05:33 PM

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#29
Aug1010, 09:13 AM

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#30
Jun3012, 05:53 AM

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[itex] \vec E = \frac{\hat r}{4 \pi r^2} \times \frac{Q}{\epsilon_0} \\
\vec E \cdot \hat r 4 \pi r^2 = \frac{Q}{\epsilon_0} \\ \text{Total Flux} = \frac{Q}{\epsilon_0} \\ \oint_s \vec E \cdot \hat n ds = \frac Q{\epsilon_0}[/itex] Which is integral form of Gauss law. 


#31
Jun3012, 09:26 AM

P: 3

I'm fairly sure this can be done with lagrangians as long as you are comfortable with the jump from lagrangians to lagrangian densities.



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